How to calculate conductivity when you know the electron and hole mobility?

Heres the question:
A sample of pure silicon, which has four valence electrons is doped with gallium, which has three valence electrons, to give a concentration of 10^23 m^-3.

a. Is the material n-type or p-type?

b. If the electron mobility is 0.14 m^2/Vs and the hole mobility is 0.05 m^2/Vs, what is the conductivity?

I know this is a p-type, just want to make sure. And the equation for conductivity is conductivity = (number denisty)*(electron charge)*(mobility), but I don't know how to incorporate electron and hole mobility into this.

An explanation and working out on how you solve this question especially part b) would really be helpful. Thanks.

Ok thanks so concentration of electrons(n) is the number of electrons per unit volume and the concentration of holes(p) is the number of holes per unit volume. What I don't get is how would you find these quantities, are they specific to the type of material for e.g. in this case it is silicon doped with gallium. I found this formula for "n" which was n=N/V where N is the number of objects in volume V, that didn't help much though :S.

I just need to know how or where do you find the values for 'n' and 'p' in this question.

Heres the question:
A sample of pure silicon, which has four valence electrons is doped with gallium, which has three valence electrons, to give a concentration of 10^23 m^-3.

a. Is the material n-type or p-type?

b. If the electron mobility is 0.14 m^2/Vs and the hole mobility is 0.05 m^2/Vs, what is the conductivity?

we agreed that part(a) was fine ..

I will try to explain part(b) in a clearer way ..

lets try to analyze this problem in a logical way, at the beginning we had a sample of silicon with an n (concentration of electrons) and p (concentration of donors) and that sample has no net charge ..

then, we doped Ga atoms , and since Ga is from group 3 and Si from group 4, then Ga atoms are considered as acceptors to Si (I assume you understand this part) .. since we doped atoms (not ions) then the Si sample will still be neutral (that is why i suggested to use the neutrality equation) ..

now lets get back to what part(b) is asking, you are asked to find the conductivity, and as I mentioned in a previous post:

conductivity = q(n*Mn + p*Mp) .. >> this is the general formula

you are given Mn (electron mobility), Mp (hole mobility), and q is just 1.6x10^-19c, so in order to get your answer you still need to know what are n and p ..

two unknowns need two equations (which I suggested for you to use)

one equation can be used (the neutrality equation) :

n+Na = p+Nd , your sample is just doped with acceptor so Nd=0

and you know that ni^2 = n*p, (I think ni should be given somewhere in your book as constant, I believe that ni = 1.5x10^10 cm-3 << check this value againa)

I tried to explain it in a simple way as much as I could, it is straightforward now solve the two equations to get your answer!!! I can not do this for you!

I have exactly the same question and I tried to solve it using simultaneous equations you suggested, but I ended up with a stupid quadratic and two impossible answers - that the value of n is 0 or negative....

Also, we haven't been taught most of the formulas you're using so I'm pretty sure we shouldn't have to use them. We only know the first one...

Furthermore, the semi-conductor is a p-type, extrinsic semi-conductor, so why are we using a formula that we were told is for intrinsic semi-conductors...? The provision of hole and electron mobility could be to stop people from guessing the first answer from the info in the second part - ie, if you only had hole mobility, you'd know to use that formula - p*q*Mp, right?

Lastly, WTH does the "concentration of 10^23" refer to? Electron concentration, gallium concentration or what? Because I'll bet it's useful in some way and you haven't even suggested a use for it... Unfortunately the stupid question isn't clear about what it is either... if it is the concentration of gallium, how can we figure out n or p from it?

hmm, I thought I made this question clearer (but it seems Im terrible in explaining :s) ..

(>> taking a deep breath) I will try again .. first, I will answer your questions:

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(I am not sure what equations they have been teaching you, so the only way to answer your questions is to use what I know)
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the semi-conductor is a p-type, extrinsic semi-conductor, so why are we using a formula that we were told is for intrinsic semi-conductors...?

I believe that you are refering to the equation ni^2 = n*p .. well to be more accurate it is ni^2 = no*po (where no and po: are the concentrations of electrons and holes, respectively, provided thermal equilibrium is maintained) this equation is valid for intrinsic or doped material ..(why is that true?)

The provision of hole and electron mobility could be to stop people from guessing the first answer from the info in the second part - ie, if you only had hole mobility, you'd know to use that formula - p*q*Mp, right?

I woud say that it is not necessary true all the time, as I have mentioned the general formula for conductivity:

conductivity = q(n*Mn + p*Mp)

you can use conductivity = q*p*Mp if you know for sure that the value of q*n*Mn can be neglected when compared with q*p*Mp, and you can not be sure unless you found out what are n and p ..

I hope I answered all your questions regarding this problem, if there are equations you are not familiar with please mention what exactly are they, also it would be better if you write all the equations related to the problem so I would be able to help..

well, I have in my book example of doped Si and they were using the relation ni^2=n*p ..

Anyway, you can make an approximation that Na = p if Na is much greater than ni ..

And as I mentioned in a previous post, the conductivity in general = qnMn+qpMp (intrinsic or extrinsic) you can not ignore one term unless you know for sure that it would be too small compared to the other one, sometimes you would have a sample doped (say with acceptor) but the doped concentration is not large enough, so in the case you can not say that the conductivity is just qpMp, you got my point there?

bigstar25 - Thanks heaps for your help and although you may be right about the use of that formula for extrin. and intrin. we (being first years) were only told to use it for intrin. NOT extrin. so I and the other students I checked with just used the conductivity formula with hole mobility only. Also, our newbie first year books haven't told us about ni^2 = np thing yet :)

newbiephysics - you're right - that concentration was for the acceptor and turns out we were supposed to assume that the concentration of Ga = p since each atom of Ga has one 'hole' spot available.

hmm, well, I did mentioned in a previous post that the number 10^23 was most likely for the Ga atoms (concentration of acceptors) , also I said in case this Na much larger than ni (which it is in this case =1.5x10^16 m-3) then Na=p ...

For the conductivity in this question the dominant term is as you said q*p*Mp .. And I still insist that these kind of approximations can not be applied all the time (just keep that on mind in case you will take an advanced class) ..

And I appologize in case I confused you, but I believe I didnt mention wrong information..

Hmm I was just following the whole thread and wanted to mention that thebigstar25 is right.

p*n=ni^2 formula is valid for both intrinsic and extrinsic conductors.

The problem is when we try to solve the doped semiconductor conductivity problem, we always assume that Nd+ (ionized portion of dopants) >>Na_ or Na_>>Nd+. Hence say if the semiconductor is p type, p>>n, so we cancel n on the RHS of Nd+ + p=Na_ + n and Nd+ on the LHS and use the new equality p=Na_. For general cases where p and n values are comparable, we need to solve p or n values numerically. Therefore we only consider either p>>n or n>>p cases to solve for the conductivity by hand.

If you wonder, my reference is Kittel's Solid State Physics edition 3.